Question

for the interval [0,2pi) find the values of x such that cos2x=-sin^2x

Answer 1

cos2x=-sin^2x
1=-tan2x
tan2x=-1

Answer 2

@.Sam. idk how u got -1

Answer 3

help!

Answer 4

@.Sam. can u please explain how u got that?

Answer 5

1=-tan2x
multiply -1 to both sides,
tan2x=-1

Answer 6

@.Sam. how did u get tan2x

Answer 7

divide both sides by cos2x

then \[\frac{\sin2x}{\cos2x}=\tan2x\]

Answer 8

move cos2x to the other side, so wo can get 1=-tan2x
tan2x=-1 we know that tan(5pi/4)=-1
so x=5pi/8

Answer 9

@AndrewNJ i thought -1 was the answer..so how did u get 5pi/8

Answer 10

tan2x=-1
No this is not the final answer, u need to use the chart
\[2x=\tan^{-1}(-1)\]